Congruency of Triangles – Class VIII

Congruency is one of the fundamental concepts in geometry. This concept is used to classify the geometrical figures on the basis of their shapes. Two geometrical figures are said to be congruent, if they have same shape and size. For example:

Two line segments are congruent if they have same length.

Two angles are congruent if they have same measures.

Two circles are congruent if they have same radii.

Two squares are congruent if they have sides of same measures.

Congruency of Triangles:

Two triangles are congruent if all the angles and sides of one triangle are equal to the corresponding angles and sides of the other triangle.

A few words about the use fo this notation. When we write ∆ABC ≅ ∆DEF, the important things is to observe that the vertices A, B and C corresponding to the vertices D, E and F in that order. If we write ∆ABC ≅ ∆EFD, this gives a different meaning. This means AB = EF, BC = FD and CA = DE; ∠A = ∠E, ∠B = ∠F and ∠C = ∠D.

Corresponding sides and angles – Congruency of Triangles

Let us say that, on superscription, triangle ABC covers triangle DEF exactly in such way that,

AB = DE, AC = DF and BC = EF;

∠A = ∠D, ∠B = ∠E and ∠C = ∠F.

Then, ∆ABC ≅ ∆DEF

Angles which coincide on superscription are called corresponding angles. Sides which coincide on superscription are called corresponding sides.

Congruency of Triangles – Exercise 3.3.1

Identify the corresponding sides and corresponding angles in the following congruent triangles.

Solution:

PQ = XY, PR = XZ , QR = YZ

(ii)

PR = AC, QP = AB

2. Pair of congruent triangles and incomplete statements related to them are given below. Observe the figures carefully and fill up the blanks.

5. Suppose ABC is an equiangular triangle. Prove that it is equilateral. (You have seen earlier that an equilateral triangle is equiangular. Thus for triangles triangularity is equivalent to equilaterality)
Solution :

3. In the figure, two sides AB, BC and the median AD of Δ ABC are respectively equal to two sides PQ, OR and median PS of Δ PQR. Prove that (i) Δ ABD ≅ ΔPSQ; (ii) Δ ADC ≅ Δ PSR. Does it follow that triangle ABC and PQR are congruent?

Some consequences – Congruency of triangles

Proposition 1: Suppose two sides of a triangle are not equal. Then the angle opposite to a larger side is greater than the angle opposite to the smaller side.

Given: A triangle ABC in which AC > AB.

To prove: ∠B > ∠C

Construction: take apoint D on AC such that AB = AD. (this is possible since AC > AB)

Proof: In triangle ABD, we have AB = AD(by construction)

Therefore ∠ABD = ∠ADB (angles opposite to equal sides)

Now, ∠BDC is an exterior angle for triangle BCD. Hence it is larger than interior angle ∠BCD. We thus get,

∠C > ∠BDA = ∠ABD < ∠ABC = ∠B

Proposition 2: In a triangle, if two angles are unequal, then the side opposite to the larger angle is greater than the side opposite to the smaller angle.

Given: A triangle ABC in which ∠B > ∠C.

To prove: AC > AB

Proof: Observe that, ∠B > ∠C Therefore, AC ≠ AB.

For, AC = AB implies that ∠B = ∠C

Thus either,

AC < AB or AC > AB

IF AC > AB, then by Proposition 1: Suppose two sides of a triangle are not equal. Then the angle opposite to a larger side is greater than the angle opposite to the smaller side. ∠B > ∠C, but; this contradicts the given hypothesis. The only possibility left out is AC > AB.

Proposition 3: In a triangle, the sum of any two sides is greater than the third side.

Given: A triangle ABC

To prove: AB + AC > BC

Construction: Extend BA to D such that AD = AC and join DC.

Proof:

Then, BD = BA + AD = BA + AC

Since AD = AC, we have,

∠ADC = ∠ACD(angle opposite to equal side)

Hence we obtain, ∠BCD > ∠ACD = ∠ADC = ∠BDC

In triangle BCD, we have,

∠BCD > ∠ACD this implies BD > BC (by proposition 2)

But BD = BA + AC as we have observed earlier. WE thus get

BA + AC > BC

We can similarly prove CA < AB + BC and AB < BC + CA.

Congruency of triangles – Exercise 3.3.7

1. In a triangle ABC, ∟B=28° and ∟C =56°. Find the largest and the smallest sides.